Expanding the Commutator of 3 Operators A, B, and C - Quantum/Math Question

  • Thread starter Thread starter cosmic_tears
  • Start date Start date
AI Thread Summary
The discussion revolves around expanding the commutator of three operators A, B, and C, specifically [A, BC], in terms of the two-operator commutators [A, B], [A, C], and [B, C]. The user expresses frustration in trying to simplify the expression without eliminating operators, questioning the feasibility of reducing a sum of three operators to two. Clarification is sought regarding the nature of the expansion and whether it can be expressed solely in terms of the mentioned commutators. The user concludes that the task may not be possible as demonstrated by a specific example involving momentum and position operators. The conversation highlights confusion among students regarding the assignment's requirements and the complexities of operator algebra.
cosmic_tears
Messages
47
Reaction score
0
Quantum/math question :)

Homework Statement


I need to expand the commutator of the 3 operators A,B,C: [A,BC] in terms of [A,B], [A,C], [B,C].

Homework Equations


[A,B] is defined to be AB - BA
also, [A,BC] = B[A,C] +[A,B]C.
There are some other identities but none that I see relevant.



The Attempt at a Solution



Tried lots of opening and additing/substructing of elements using the above formulas but I could never get rid of "operators". Is the a way to write "B" and "C" in terms of the above?
I'm working on this for over an hour! :(

Thank you very much!
Tomer.
 
Physics news on Phys.org
Are you sure you're supposed to get rid of the operators? I don't think it's possible to express it otherwise. That would require reducing a sum of 3 operators to a sum of 2 operators.
 
Thanks for responding.

The question is:
Let A;B;C; be operators
1. Expand [A;BC] in terms of [A;B]; [A;C]; [B;C]

If you can understand anything different from the exact question you're most welcomed to explain to me :)
What do you mean by "That wuold require reducing a sum of 3 operators to a sum of 2"?
Why is that a consequence?
 
well an expansion of [A,BC] is ABC-BCA which is a sum of 3 operators in a row. I don't see how you could express that in the form aAB + bBA +cAC +dCA + eBC + fCB (where a...f are constants) which is the form it would need to take if you wanted it expressed solely in terms of [A,B], [A,C] and [B,C]
 
I still don't understand the "paradox" here, nor do I see how the sum you wrote (with the consts a...f) is a sun of two operators... what I see is 6 operators.

Ah well... I thank you for trying anyway!
 
Your answer will involve a sum of the products of the original operators with the two-operator commutators, as in:

X*[Y,Z] or [X,Y]*Z

Where X, Y, and Z are each one of A, B, and C. (I wrote it this way to avoid giving the answer away :)
 
But I need an expression consisting only of [A,B], [A,C] and [B,C]
I already know that [A,BC] = B[A,C] + [A,B]C...
 
But clearly that's not possible. For example, let

A=p_{x}, B=x, C=y

Then

[A,BC] = [p_{x},x]*y = -i \hbar *y

and

[A,B] = -i\hbar, [A,C] = [B,C] = 0

There just isn't any way to do it. I think they mean to write it as an expression where the commutators involve only two of A, B, and C, like the one you gave above.
 
*sigh*... I'm really going to hate them if you're right... :)
I'll send a mail to my tutor, all the student are too confused with this subject I guess none have noticed yet.

Thank you very much!
I'll post here if it turns out otherwise...
 
Back
Top